Let's solve the problem step-by-step: --- ## **Problem Restatement** A viscous fluid is between two infinitely long, vertical, concentric cylinders: - **Outer cylinder:** radius \( r_o \), rotates with angular velocity \( \omega \) - **Inner cylinder:** radius \( r_i \), fixed (does not rotate) **Find:** The velocity distribution \( v_\theta(r) \) in the gap, using the Navier-Stokes equations. **Assumptions:** - Flow is axisymmetric (\( \frac{\partial}{\partial \theta} = \)) - Velocity is only in the tangential (\( \theta \)) direction. - Only body force is weight (acts vertically, so does not affect tangential motion). --- ## **Step 1: Navier-Stokes Equations in Cylindrical Coordinates** The θ-component (tangential) of the steady, incompressible Navier-Stokes equation in cylindrical coordinates (with no dependence on \( z, \theta \), and only \( v_\theta \neq \)) is: \[ = \mu \left[ \frac{\partial^2 v_\theta}{\partial r^2} + \frac{1}{r} \frac{\partial v_\theta}{\partial r} - \frac{v_\theta}{r^2} \right] \] --- ## **Step 2: Simplifying the Equation** Set the left side to zero (steady-state, no pressure gradient or body force in θ): \[ \frac{\partial^2 v_\theta}{\partial r^2} + \frac{1}{r} \frac{\partial v_\theta}{\partial r} - \frac{v_\theta}{r^2} = \] --- ## **Step 3: Guess the Solution** This is a standard ODE with a general solution: \[ v_\theta(r) = A r + \frac{B}{r} \] where \( A \) and \( B \) are constants to be determined by boundary conditions. --- ## **Step 4: Apply Boundary Conditions** - At \( r = r_i \): **Inner cylinder is fixed** \[ v_\theta(r_i) = \] - At \( r = r_o \): **Outer cylinder rotates with angular velocity \( \omega \)** \[ v_\theta(r_o) = \omega r_o \] Plug these into the general solution: ### **At \( r_i \):** \[ = A r_i + \frac{B}{r_i} \] \[ A r_i = -\frac{B}{r_i} \] \[ A = -\frac{B}{r_i^2} \] ### **At \( r_o \):** \[ \omega r_o = A r_o + \frac{B}{r_o} \] Substitute \( A \) from above: \[ \omega r_o = \left(-\frac{B}{r_i^2}\right) r_o + \frac{B}{r_o} \] \[ \omega r_o = -\frac{B r_o}{r_i^2} + \frac{B}{r_o} \] \[ \omega r_o = B \left( -\frac{r_o}{r_i^2} + \frac{1}{r_o} \right ) \] \[ \omega r_o = B \left( \frac{-r_o^2 + r_i^2}{r_i^2 r_o} \right ) \] \[ B = \frac{\omega r_o r_i^2 r_o}{r_i^2 - r_o^2} \] \[ B = \frac{\omega r_o^2 r_i^2}{r_i^2 - r_o^2} \] Now, substitute \( B \) back into \( A = -\frac{B}{r_i^2} \): \[ A = -\frac{1}{r_i^2} \cdot \frac{\omega r_o^2 r_i^2}{r_i^2 - r_o^2} = -\frac{\omega r_o^2}{r_i^2 - r_o^2} \] --- ## **Step 5: Final Velocity Profile** Plug \( A \) and \( B \) into the general solution: \[ v_\theta(r) = A r + \frac{B}{r} \] \[ = -\frac{\omega r_o^2}{r_i^2 - r_o^2} r + \frac{\omega r_o^2 r_i^2}{r_i^2 - r_o^2} \cdot \frac{1}{r} \] \[ = \frac{\omega r_o^2}{r_i^2 - r_o^2} \left( -r + \frac{r_i^2}{r} \right) \] Or, equivalently: \[ \boxed{ v_\theta(r) = \omega r_o^2 \frac{r_i^2 - r^2}{r (r_i^2 - r_o^2)} } \] --- ## **Summary** ### **Exact Solution for the Tangential Velocity:** \[ \boxed{ v_\theta(r) = \omega r_o^2 \frac{r_i^2 - r^2}{r (r_i^2 - r_o^2)} } \] where: - \( r_i \) = radius of inner (stationary) cylinder - \( r_o \) = radius of outer (rotating) cylinder - \( \omega \) = angular velocity of outer cylinder --- **This is the velocity distribution for a viscous fluid between two concentric cylinders with the inner one fixed and the outer one rotating.**